Question

Calculate the maximum kinetic energy and maximum velocity of the photoelectrons emitted when the stopping potential is 81 V for the photoelectric emission experiment.

Answer 1

V₀ = 81 V

e = 1.6 × 10⁻¹⁹ C

m = 9.1 × 10⁻³¹ kg

Maximum kinetic energy of electron,

K_max = `"eV"_0`

= 1.6 × 10⁻¹⁹ × 81

= 129.6 × 10⁻¹⁹

= 1.29 × 10⁻¹⁷

K_max = 1.3 × 10⁻¹⁷ J

Maximum velocity of photoelectron,

v_max = `sqrt((2"eV"_0)/"m")`

= `sqrt((2 xx 1.6 xx 10^-19 xx 81)/(9.1 xx 10^-31))`

= `sqrt((259.2 xx 10^-19)/(9.1 xx 10^-31))`

= `sqrt(28.48 xx 10^12)`

v_max = 5.3 × 10⁶ ms⁻¹